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Proof step 5 from proof num-i2n-ls-02 (previous step)

0 ≤ i ⊦ n→i(m) < n→i(i→n(i)) ↔ n→i(m) < i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
0 ≤ i ⊦ n→i(m) < abs(i) ↔ n→i(m) < i

Rule argument

Additional information

Used simplifier rules: ini2-abs: ⊦ n→i(i→n(i)) = abs(i);